Someone help me out. I swear I leanred how tro do this years ago but I guess I'm not that smart
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Using the t-distribution, it is found that the 99% confidence interval for the difference is (0.058, 3.462).
We are given the <u>standard deviation for the samples</u>, hence, the t-distribution is used to solve this question.
The standard errors are given by:
For the distribution of differences, the <u>mean and the standard error</u> are given by:
The interval is given by:
The critical value for a <u>two-tailed 99% confidence interval</u> with 18 + 18 - 2 = <u>34 df</u> is t = 2.7284.
Then:
The 99% confidence interval for the difference is (0.058, 3.462).
For more on the t-distribution, you can check brainly.com/question/16162795